# Flux through surface (correct or not?)

I need to calculate the flux of the vector field

$$\mathbf{v}=(x-y,\; x+2y,\; z)$$

through the surface

$$S\, : \, \mathbf{r}(u, v)=(\frac{u^2}{\sqrt{2}},\, uv,\, \frac{v^2}{\sqrt{2}}),\;\; 0≤u≤1,\; -1≤v≤1,\; \mathbf{n}\cdot \mathbf{e}_x ≥0$$

I would like to know whether or not this is correct.

$$\mathbf{n}=\frac{(\mathbf{r}_u \times \mathbf{r}_u)}{|\mathbf{r}_u \times \mathbf{r}_u|},\;\;\; dS=|\mathbf{r}_u \times \mathbf{r}_u|dudv,\;\;\; d\mathbf{S}=(\mathbf{r}_u \times \mathbf{r}_u)dudv$$

$$(\mathbf{r}_u \times \mathbf{r}_v)= \begin{bmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \sqrt{2}u & v & 0 \\ 0 & u & \sqrt{2}v \\ \end{bmatrix}=(\sqrt{2}v^2,\; -2uv ,\; \sqrt{2}u^2)$$

$$\iint\limits_S \mathbf{v}\cdot d\mathbf{S} =\iint\limits_S \mathbf{v}\cdot \mathbf{n}d\mathbf{S} =\iint\limits_S \mathbf{v}\cdot (\mathbf{r}_u \times \mathbf{r}_v)dudv =$$ $$=\iint\limits_S(x-y,\; x+2y,\; z)\cdot (\sqrt{2}v^2,\; -2uv ,\; \sqrt{2}u^2)dudv=$$ $$=\int\limits_{-1}^1 \left(\int\limits_0^1 \left( (\frac{u^2}{\sqrt{2}}-uv,\; \frac{u^2}{\sqrt{2}}+2uv,\; \frac{v^2}{\sqrt{2}})\cdot (\sqrt{2}v^2,\; -2uv ,\; \sqrt{2}u^2)\right)du\right)dv=$$ $$=\int\limits_{-1}^1 \left(\int\limits_0^1 \left((\frac{u^2}{\sqrt{2}}-uv)\sqrt{2}v^2+(\frac{u^2}{\sqrt{2}}+2uv)(-2uv)+\frac{v^2}{\sqrt{2}}\sqrt{2}u^2\right)du\right)dv=$$ $$=\int\limits_{-1}^1 \left(\int\limits_0^1 \left(u^2v^2-\sqrt{2}uv^3-\sqrt{2}u^3v-4u^2v^2+u^2v^2\right)du\right)dv=$$ $$=\int\limits_{-1}^1 \left(\int\limits_0^1 \left(-2u^2v^2-\sqrt{2}uv^3-\sqrt{2}u^3v\right)du\right)dv=$$ $$=-\int\limits_{-1}^1 \left(\int\limits_0^1 \left(2u^2v^2+\sqrt{2}uv^3+\sqrt{2}u^3v\right)du\right)dv=$$ $$=-\int\limits_{-1}^1 \left( \Big[\frac{2}{3}u^3v^2+\frac{1}{\sqrt{2}}u^2v^3+\frac{1}{2\sqrt{2}}u^4v \Big]_{u=0}^{u=1} \right)dv=$$ $$=-\int\limits_{-1}^1 \left(\frac{2}{3}v^2+\frac{1}{\sqrt{2}}v^3+\frac{1}{2\sqrt{2}}v\right)dv=$$ $$=-\Big[ \frac{2}{9}v^3+\frac{1}{4\sqrt{2}}v^4+\frac{1}{4\sqrt{2}}v^2 \Big]_{-1}^1 =$$ $$=-\frac{4}{9}$$

• It looks completely correct...but the final result seems to be $\;\color{red}-\frac49\;$ ... – DonAntonio Mar 19 '18 at 8:26
• @DonAntonio Oh, yeah, of course! Thanks! – Filip Mar 19 '18 at 8:28